Self-inductance of solenoid

Consider a solenoid of length l, no. of turns N, n = N/l, area A, and negligible resistance. In its middle is placed a single-turn loop of same area (well, slightly smaller) in the solenoid's middle. The loop has, for the moment, a very high resistance so there is no appreciable loop current. The solenoid is sinusoidally excited with voltage v1. ø11 is the flux in the solenoid due to the current i1 in the solenoid, and ø21 is the flux in the loop generated by the current i1 in the solenoid. Obviously, ø21 = ø11.

So this would seem a tightly coupled ideal transformer since all the flux generated by the solenoid also cuts the area of the loop, so there is no leakage flux nor turns resistances (the loop resistance forms the load for the transformer & so is external to the transformer).

Also, M = k√(L1 L2) where L2 is the self-inductance of the loop. Since the system is tightly coupled, k ~ 1.

This then says L2 = M2/L1 = μ0A/l
but this cannot be since it says L2 is a function of a property of the solenoid (namely l). In fact, computing the self-inductance of the loop is known to be exceedingly difficult, requiring elliptical integrals and what-not.

Now, an objection could be raised: since there is negligible current in the loop, the loop's self-inductance is meaningless. So let's now impose finite resistance R to the loop:

This is then a transformer with a secondary load R and secondary (loop) current i2. The equations for this transformer, assuming zero solenoid resistance, are

Why does the difficulty of a computation matter to a result? "Nature" comes up with results which otherwise require difficult computations all the time ...

This completely misses my point, which is that the self-inductance of the loop computes two entirely disparate ways, giving two entirely disparate answers. In the case of the loop within the solenoid it turns out to be a simple function of its (the loop's) area and the length of the solenoid, which makes no sense.

Whereas the stand-alone computation of the loop inductance from basics is extremely complex, is not a function of its presence in an inductively coupled system, and in fact is a function not only of its area but even the radius of the loop wire.

This should give you an idea of the complexity of computing the loop self-inductance:

Consider a solenoid of length l, no. of turns N, n = N/l, area A, and negligible resistance. In its middle is placed a single-turn loop of same area (well, slightly smaller) in the solenoid's middle. The loop has, for the moment, a very high resistance so there is no appreciable loop current. The solenoid is sinusoidally excited with voltage v1. ø11 is the flux in the solenoid due to the current i1 in the solenoid, and ø21 is the flux in the loop generated by the current i1 in the solenoid. Obviously, ø21 = ø11.

So this would seem a tightly coupled ideal transformer since all the flux generated by the solenoid also cuts the area of the loop, so there is no leakage flux nor turns resistances (the loop resistance forms the load for the transformer & so is external to the transformer).

Also, M = k√(L1 L2) where L2 is the self-inductance of the loop. Since the system is tightly coupled, k ~ 1.

This then says L2 = M2/L1 = μ0A/l
but this cannot be since it says L2 is a function of a property of the solenoid (namely l). In fact, computing the self-inductance of the loop is known to be exceedingly difficult, requiring elliptical integrals and what-not.

Now, an objection could be raised: since there is negligible current in the loop, the loop's self-inductance is meaningless. So let's now impose finite resistance R to the loop:

This is then a transformer with a secondary load R and secondary (loop) current i2. The equations for this transformer, assuming zero solenoid resistance, are

1. The formula for mutual inductance M = k√(L1 L2) holds if and only if, when calculating the separate inductances as if the other coil were absent, the magnetic path is the same for both coils.

This obviously obtains for example in a toroid transformer but not in the case I describe above.

2. Similarly, if a sinusoidal primary voltage, perfect coupling, and a real secondary load are assumed, there is no imaginary component in the secondary voltage if and only if the condition in (1) obtains. Alternatively said, a step input of voltage gives a step output of voltage if and only if the above-said magnetic paths are identical.